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Using Number set 39, 78, 321, 145 convert into ...
Hexadecimal Representation
3916Note:∴7816Note:∴32116Note:∴14516Note:∴392161×2+160×732+7=3939 in hexadecimal form is 27784161×4+160×1464+14=7878 in hexadecimal form is 4E321201162×1+161×4+160×1256+64+1=321321 in hexadecimal form is 1411459161×9+160×1144+1=145145 in hexadecimal form is 91÷16remainder7÷16remainder2÷16remainder14orE÷16remainder4÷16remainder1÷16remainder4÷16remainder1÷16remainder1÷16remainder9
Binary Representation
392Note:∴782Note:∴3212Note:∴1452Note:∴3919942125+22+21+2032+4+2+1=3939 in binary form is 10 011178Same as before question3919942125+22+21+2032+4+2+1=7878 in binary form is 100 11103211608040201052128+26+20256+64+1=321321 in binary form is 1 0100 0001145723618942127+24+20128+16+1=145145 in binary form is 1001 0001÷2remainder1÷2remainder1÷2remainder1÷2remainder0÷2remainder0÷2remainder1÷2remainder0÷2remainder1÷2remainder1÷2remainder1÷2remainder0÷2remainder0÷2remainder1÷2remainder1÷2remainder0÷2remainder0÷2remainder0÷2remainder0÷2remainder0÷2remainder1÷2remainder0÷2remainder1÷2remainder1÷2remainder0÷2remainder0÷2remainder0÷2remainder1÷2remainder0÷2remainder0÷2remainder1
Using Number Set #2 (four binary numbers) 11101101b, 00011011b, 11110101b, 01100100b, show the value of each number as:
An unsigned decimal value
A signed decimal value
271282664253224162382242122011110 1101 b127126125024123122021120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in unsigned decimal form is 23702702602512402302212112016 + 2 + 1 = 19∴1110 1101 in signed decimal form is -190001 1011 b02702602512412302212112016 + 8 + 2 + 1 = 27∴0001 1011 in unsigned decimal form is 27127126125024123122021120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in signed decimal form is -2371111 0101 b127126125124023122021120128 + 64 + 32 + 16 + 4 + 1 = 245∴1111 0101 in unsigned decimal form is 2450270260250241230221211208 + 2 + 1 = 11∴1111 0101 in signed decimal form is -110110 0100 b02712612502402312202102064 + 32 + 4 = 100∴0110 0100 in unsigned decimal form is 100127026025124123022121120128 + 16 + 8 + 2 + 1 = 155∴0110 0100in signed decimal form is -155
[10] Using Number Set #2 (four binary numbers) 01100100b, 10011010b, 01101101b, 11000110b, where the first number is considered to be A, the second number B, the third C, etc… calculate each of the following:
A v B, A v C, A v D ( v is the symbol for the OR operation)
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A ^ B, A ^ C, A ⊕ D (^ is the symbol for the AND operation, ⊕ is the symbol for the XOR operation)
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Using Number Set #2 (four binary numbers) 11101101b, 00011011b, 11110101b, 01100100b, show the value of each number as:
An unsigned decimal value
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A signed decimal value
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271282664253224162382242122011110 1101 b127126125024123122021120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in unsigned decimal form is 23702702602512402302212112016 + 2 + 1 = 19∴1110 1101 in signed decimal form is -190001 1011 b02702602512412302212112016 + 8 + 2 + 1 = 27∴0001 1011 in unsigned decimal form is 27127126125024123122021120128 + 64 + 32 + 8 + 4 + 1 = 237∴1110 1101 in signed decimal form is -2371111 0101 b127126125124023122021120128 + 64 + 32 + 16 + 4 + 1 = 245∴1111 0101 in unsigned decimal form is 2450270260250241230221211208 + 2 + 1 = 11∴1111 0101 in signed decimal form is -110110 0100 b02712612502402312202102064 + 32 + 4 = 100∴0110 0100 in unsigned decimal form is 100127026025124123022121120128 + 16 + 8 + 2 + 1 = 155∴0110 0100in signed decimal form is -155
Using Number Set #2 (four binary numbers) 01100100b, 10011010b, 01101101b, 11000110b, where the first number is considered to be A, the second number B, the third C, etc… calculate each of the following:
A v B, A v C, A v D ( v is the symbol for the OR operation)
A ^ B, A ^ C, A ⊕ D (^ is the symbol for the AND operation, ⊕ is the symbol for the XOR operation)
Using Number Set#3 - HEX 71AF2523h, 2B988398h, 9E5E4AD8h, 6B7C3487h (four hexadecimal numbers) where the first number is Q, the second number R, the third S, etc… calculate each of the following:
Q + R, Q + S, Q + T (show the carry value – 9th digit, if there is a carry value)
There is a carry for the 9th digit for Q + S
Q – R, Q – S, Q – T (use the 16’s compliment method)
111F252371AF2523+2B9883989D47A8BB71AF2523−2B98839816’s complement of RFFFFFFFF−2B988398+2B988391D4677C68101111000D4677C68+71AF25234616A18B1111F252371AF2523+9E5E4AD8100D6FFB71AF2523−9E5E4AD816’s complement of SFFFFFFFF−9E5E4AD8+2B98839161A1B5280111000061A1B528+71AF2523D350DA4A111F252371AF2523+6B7C3487DD2B59AA71AF2523−6B7C348716’s complement of TFFFFFFFF−6B7C3487+2B9883919483CB791011110009483CB79+71AF25230632F09CQ:71AF2523hR:2B988398hS:9E5E4AD8hT:6B7C3487h